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Thermochemistry

Energy:The capacity to do work or to produce heat

★ The law of conservation of energyEnergy can be convertedbut the total is a constant

Two types of energy:Kinetic energy — due to motion

½ mv2Potential energy — due to position, composition,

attraction, repulsion…….

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A

B A

B

Example

Potential energy has changed for A and B:A: decreasedB: increasedFrictional energy is involved

Total energy : no change

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◎ Energy transfer: through work and heat

Work: a force acting over a distance

In previous example: work is done on B by A

For A: different path different heat and workthe change of PE for A the same

A

B A

B

◎ The nature of energy

★ Energy is a state function — independent of path

Heat and work are path dependent

NOT state functionFrom one state to another state

may go through different path

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CH4(g) + 2O2(g)

CO2(g) + 2H2O(g)

E

PE: released as heat

※ Chemical energy

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + heat

The PE stored in chemical bonds is released to the surroundings

An exothermic reaction

Energy diagram

Ex:

N2(g) + O2(g)

2NO(g)PE

PE: flows into the system

N2(g) + O2(g) + heat 2NO(g)

An endothermic reaction

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※ Thermodynamics

The study of energy and its interconversions

Questions:How far will a reaction proceed?How much energy will be released or consumed?

★ The first law of thermodynamicsThe law of conservation of energy

※ Internal energy (E)

Define internal energy E:the sum of PE and KE in the system

Difficult to measure

E = q + w

heatwork

The change is measurable:

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◎ The sign of q and w

From the system’s point of view:

q absorbed by the system is positive

Increases E of the system

q released by the system is negative

Decreases E of the system

Ex: endothermic reaction: q (＋)exothermic reaction: q (－)

w done by the system is negative

Decreases E of the system

w done by the surroundings on the system is positive

Increases E of the system

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◎ A model study

｜work｜=｜force × distance︱=｜F × h ｜

Since P = F/A F = PA

｜work｜=｜P × A × h︱=｜P × V｜

idealgas

expansion

against PP =

FA

P

h

Work is done by the system negative

V = Vfinal － Vinitial

Vfinal > Vinitial (for expansion)V positive (for expansion)

Therefore we must definew = PV

For compressionV negativew positive

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Ex: A balloon

4.00 × 106 L 4.50 × 106 Lheat

1.3 × 108 J

Expands against 1.0 atm, E?

Soln: E = q + w

q = +1.3 × 108 Jw = PV

= (1.0 atm)(0.50 × 106 L)= 5.0 × 105 atm．L= 5.0 × 105)(101.3) = 5.1 × 107 J

1 atm‧L = 101.3 J

E = +(1.3 × 108) – 5.1 × 107 J= 8 × 107 J

※ Enthalpy (焓) and calorimetry (量熱法)

◎ The definition of enthalpy

P

higher T lower T

P

E = q + w= q – PV

At constant P: E = qP – (PV)

qP = E + (PV)= Ef – Ei + (PV)f – (PV)i= Ef + (PV)f – Ei – (PV)i

Define enthalpy H = E + PV Hf = Ef + (PV)f Hi = Ei + (PV)i qp = Hf – Hi = H

★ qP = H at constant P= E + PV at constant P qP ≠ E (unless V = 0)

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● What’s so important?Usually reactions are performed at constant PqP is obtained experimentallyTherefore, H can be measured directly

● The beauty of H

H = E + PV

A state function Path independent

For a reaction H = Hproducts – HreactantsExothermic reaction: qP is negative negative H

Endothermic reaction: qP is positive positive H

● H and E may be different

◎ Calorimetry (量熱學)The science of measuring heat observing T change

C =heat absorbed

increase in T

Heat capacity熱容

Specific heat capacity: per gram sampleJ

oC‧gor J

K‧g

Molar heat capacity: per mole sampleJ

oC‧molor J

K‧mol

Unit:

Unit:

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◎ Constant P calorimetry

Purpose: get qP qP = H

Ex: Mix 50.0 mL 1.0 M HCl at 25.0 oC50.0 mL 1.0 M NaOH at 25.0 oC

31.9 oC

Soln: T = 31.9 – 25.0 = 6.9 oCmass ≈ 100.0 mL × 1.0 g/mL = 1.0 × 102 g

Energy released = s × m × T= (4.18 J/oCg)(1.0 × 102 g)(6.9 oC)= 2.9 × 103 J

specific heat capacitymass

Heat is an extensive (外延) propertyrelated to the amount of substance

T is an intensive (內涵) propertynot related to the amount of substance

50.0 mL 1.0 M HCl 5.0 × 102 mol H+

2.9 × 103 J5.0 × 102 mol

= 5.8 × 104 J/mol

H+(aq) + OH(aq) H2O(l) H = 58 kJ/mol

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◎ Constant V calorimetry

V = 0 w = 0

E = qV + w = qV

Therefore, E = qV at constant V

Use a bomb calorimeterinsulating container

water

steel bomb

Ex: Combustion of 0.5269 g octane (C8H18)The bomb calorimeter used: C = 11.3 kJ/oC

T = 2.25 oC

Soln: Heat released = CT = 11.3 × 2.25 = 25.4 kJ

MW of octane = 114.2 g/mol

0.5269 g octane 0.5269 g

114.2 g/mol= 4.614 × 103 mol

For 1 mol octane:

Heat =4.614 × 103

25.4= 5.50 × 103 kJ/mol

E = 5.50 × 103 kJmol1

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※ Hess’s law

Reactants Products H = ΣnpHproducts – ΣnrHreactants

R P

B

A

∵ H is a state function∴ HRP is the same for different pathways

Ex: N2(g) + 2O2(g) 2NO2(g) H1 = 68 kJ

N2(g),2O2(g)

O2(g),2NO(g)

2NO2(g)

H2= 180

H3= 112

H1

H(kJ)

A

B

C

C

B

AH1

H2 H3

H1 = H2 + H3 180 – 112 must be 68 kJ

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※ Characteristics of H

1. H = HreverseN2(g) + O2(g) 2NO2(g) H2 = +180 kJ

endothermic

2NO2(g) N2(g) + O2(g) H2 = 180 kJ

exothermic

Reverse reaction:

2. H is an extensive propertydepending on the amount of reactants

Xe(g) + 2F2(g) XeF4(s) H = 251 kJ2Xe(g) + 4F2(g) 2XeF4(s) H = 2(251 kJ)

= 502 kJ

Ex: 2B(s) + 3H2(g) B2H6(g) H = ?Data available:(a) 2B(s) + 3/2O2(g) B2O3(s) H = 1273 kJ(b) B2H6(g) + 3O2(g) B2O3(s) + 3H2O(g) H = 2035 kJ(c) H2(g) + 1/2O2(g) H2O(l) H = 286 kJ(d) H2O(l) H2O(g) H = 44 kJ

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Soln:+a (for B)2B(s) + 3/2O2(g) B2O3(s) H = 1273 kJ

+3c (for H2)3H2(g) + 3/2O2(g) 3H2O(l) H = 286 × 3 kJ

b (for B2H6)B2O3(s) + 3H2O(g) B2H6(g) + 3O2(g) H = 2035 kJ

+3d (for H2O)3H2O(l) 3H2O(g) H = 44 × 3 kJ

2B(s) + 3H2(g) B2H6(g)H = –1273 – (286 × 3) + 2035 + (44 × 3)

= +36 kJ

Ex: 2B(s) + 3H2(g) B2H6(g) H = ?

※ Standard enthalpies of formation

Real interest: calculate H for reactionsProblem: no way to know the exact value of enthalpy

Solution: resolve each compound into the basic elements

Define a standard enthalpy of formation (Hfo)— 1 mol of a compound from its elements in their

standard states

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Standard state: The physical state in which the substance is most stable at 1 atm and the temperature of interest (usually 25 oC)

gas: a pressure of 1 atmsolution: a concentration of 1 Ma pure liquid or solid compound: the pure liquid or solidelement: the state at 1 atm (25 oC)

Ex: Na(s), Hg(l)

*Do not confused with STP

Ex: 1/2N2(g) + O2(g) NO2(g) Hfo = 34 kJ/mol

Elements in standard state

1 mol(in standard state)

C(s) + 2H2(g) + 1/2O2(g) CH3OH(l) Hfo = 239 kJ/molgraphite

◎ The use of Hfo

Ex: CH4(g) + 2O2(g) CO2(g) + 2H2O(l) Ho = ?

Data available:C(s) + 2H2(g) CH4(g) Hfo = 75 kJ/molC(s) + O2(g) CO2(g) Hfo = 394 kJ/molH2(g) + 1/2O2(g) H2O(l) Hfo = 286 kJ/mol

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

C(s)2H2(g)

C(s)O2(g)

2H2(g)O2(g)

2O2(g)

Ho = 75 + 0 – 394 – (286 × 2)= –Hfo(CH4) + Hfo(CO2) + 2(Hfo(H2O)) = Hfo(CO2) + 2(Hfo(H2O)) – Hfo(CH4)

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Conclusion:

Hrxno = ΣnpHf

o(products) – ΣnrHf

o(reactants)

(elements in their standard state are not included)

Ex: 4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(l)

Hfo : 46 34 286 kJ/mol

Ho = 6(286) + 4(34) – 4(46) = 1396 kJ

Cf. H = ΣnpHproducts – ΣnrHreactants